







LIBRARY OF CONGRESS. 




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— -^1^41 

;S OF AMEEI€A. 



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Zc^t JSook 



Mechanical Drawing, 



BEING 



An Explanation of the Principles of Geometry and Ortho= 

graphic Projection, the Helix, and Toothed Gearing as 

they are applied by Mechanical Draughtsmen, 

with Rules for Screw Cutting. 



Compiled witb ©rtoinal irilustrattons 

John S. Rooke, 

Teacher of Metal Working and Mechanical Drawing in the Workshop Schools of the 

Spring Garden Institute, Philadelphia. 



PRINTED FOR THE AUTHOR. 




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Copyrighted, 1895, by JOHN S. ROOKE. 



SPANGLER & DAVIS, PRS., PHILA 



Introduction. 



The purpose of this book is to aid pupils in the study of the principles underlying 
mechanical drawing. It is not intended to make the study easy, for the pupil cannot 
be benefited in that way. For that reason, though the drawings are made sufficiently 
accurate to serve as illustrations of the text, they are not so accurate or complete as 
to permit their use as flat studies to be merely copied without disclosing the copyist's 
ignorance of the principles intended to be taught. Nor are the explanations made so 
full and complete as to relieve the pupil of the necessity of thinking. In short, this 
-book by its text and illustrations, will enable the earnest pupil to study the art and get 
a firm grasp upon the principles of projection, but it will be of no use to the careless 
pupil who seeks to skim over his studies and make nothing more than a show of 
understanding. 

The subject is presented in a way that is likely to be interesting, and the problems 

set are of great practical value. The study is not an easy one— if it were it would be 

of less value than it is — but those who give it serious attention will soon find that 

what at first seemed hard to understand has become quite simple, and that they have 

been prepared by a thorough mastery of principles to understand and make the most 

complex mechanical drawings. This is the end in view. Copying from flat drawings 

is useful only as an exercise with the pen and other instruments ; the mastery of the 

principles of projection gives the pupil the key to original work. 

John S. Rooke. 
Philadelphia , August^ i8g^. 



Orthographic Projection. 

Our first illustration is intended to show what is meant by orthographic projection ; 
the pupil will hereafter have to imagine these planes represented by glass. 

In fig. h we have a rectangular prism placed within a case of plates of glass upon 
which the projections of the prism are made. These plates of glass represent the 
planes of projection, and can be revolved about the axes a b and b c, until all are in 
one plane as in fig. b, which is called the plane of the paper. h*" is called the hori- 
zontal plane, a"' the front vertical, and c'' the side vertical. Suppose C to be revolved 
about b c until it is in the same plane as a'', and that C and sC are revolved about 
a b until they are in the same plane as h^, then we have fig. b. 

The projections from all points of an object perpendicular to these planes of 
projection are called orthographic projections. The projection on h^' is called the top 
or plan, on a'' the front elevation, and on c'' the side elevation. It can readily be seen 
without any reference to the planes that these views are arranged as common sense 
would suggest, bringing the top to the top, the front to the front, and the side to 
the side. 

In the illustrations the small letters h and v are used with all other letters to 
indicate in one case the horizontal and in the other the vertical planes. 



General 1 nstruction. 

The drawings should be made on sheets of paper 17 x 22 inches with margin 
lines yz inch from the edge of the paper, which will make the measurement inside the 
margin lines 16 x 21 inches. The paper should be fastened on the drawing board 
with thumb tacks. One in each corner will be sufficient, care being taken to have 
the paper put on the board square, and as tightly stretched as possible. ■ 

The T square should be used with the head against the left edge of the drawing 
board (unless the person is left-handed) and the pencil be sharpened to a flat point, 
similar to a thin wedge, and always kept sharp. The pencil for the compass* must be 
sharpened in the same way. 

The drawing and compass pens must also be kept sharp and perfectly clean. 
Never lay them down or put them away without first cleaning them. A small piece 
of chamois is very good for this purpose. 

All horizontal lines should be drawn with the T square, and all vertical lines 
with the triangle set against the upper edge of the T square. 

All drawings should be made in pencil first, with as much care as though they 
were not to be inked. If a drawing is not penciled accurately and neatly, it cannot 
be expected to be neat after it is inked. Do not leave superfluous lead pencil lines on 
the drawing, but erase all except those to be inked. 

The drawing having been finished in pencil, proceed to ink in. Fill the drawing 
pen with ink ; then on a small piece of paper the same as that used for the drawing, 
try the pen until it makes such a line as is desirable for the work. Then with the 
straight edge about ^^2^ inch from the pencil line on the drawing, proceed with the 
inking, holding the pen as nearly vertical as possible. The pen should always be 
tested on the extra piece of paper, after filling, before it is used on the drawing. 

All lines forming the object should be black ; all construction and dimension 
lines, fine red. These red lines are shown on the plates by two dots and one dash, 
but orf the drawings they are to be full lines. The centre lines shouM be fine blue 
lines. They are shown on the plates by one dot and dash, but on the drawings are 
to be full lines. Any part of the object outside of a cutting plane will be drawn with 
a long and short black dot. 

All arcs should be inked first, as it is easier to draw a straight line to an arc, 
than to draw an arc to a straight line. 

All figures and arrow points should be made in black ink with a fine writing pen. 

The shape of a draughtsman's scale is such that when it is laid on the paper, the 
lines forming the divisions of the scale will come down to the paper, which enables the 
draughtsman to mark the dimensions directly from the scale. Dimensions may thus 
be marked off more quickly and accurately than by setting the dividers to the scale, 
and besides that the scale is thus preserved from being injured by the sharp points of 
the instruments. 



Plate 1. 

Fig. 1 represents full black lines to be drawn }4r inch apart. The pupil will be 
careful to draw them off accurately and ink in carefully. 

Pig. 2 represents full black lines to be drawn to the measurements given, using 
care to have them accurate. 

.Fig. 3 represents lines which the pupil will lay off to the measurements given. 
The first three lines represent a fine black line as used in drawing. The next three 
lines show how fine a red or blue line should be drawn. The first and last of the 
three should be drawn in red, and the middle one in blue. The next three lines 
represent shade lines, and should be drawn in black. The next three represent a 
fine black dotted line. The next three represent long and short dotted lines, and will 
be used to indicate parts of a figure which are cut off. 

Fig. 4 is an exercise in section lining to represent cast iron, and is to be drawn 
in black. 

Fig. 5 represents wrought iron. The section lining is to be in blue, three lines 
being drawn then a space left, followed by three more lines and a space, and so on 
until the drawing has been finished. 

Fig. 6 represents steel. It is to be in blue, and drawn like fig. 5, except that the 
middle line should be dotted. 

Fig. 7 represents brass, and the section lining is to be done in red, with two 
lines instead of three, and a space. 

The lines forming the outlines of all of these figures will be in black ink. 







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Plate 2. 

Fig. 8. To bisect a straight line. Let a b be the straight line ; then with a 
and b as centres, and any radius greater than one-half of a b, draw the two arcs c and 
d. A line through c d will bisect a b. 

Fig. 9. To bisect an arc of a circle a b c. Join a c with a straight line, and 
proceed as in fig. 8. 

Fig. 10. To draw a tangent to the arc of a circle when the centre is not 
accessible. Let a be the point on the given arc d a c. Lay off equal distances upon 
the arc from a to c and to d. Join c and d. From d as a centre, and radius d a draw 
arc a g ; then with same radius and a as a centre, draw arcs d b, c e. Then make 
d b and c e equal in length to g a. A straight line through the points thus found 
will be tangent to a." 

Fig. 11. To erect a perpendicular to a line a b from a point a or near its end. 
With a as a centre and any radius, draw the arc d c. With d as a centre and same 
radius cut d c in c. With c as a centre and the same radius, draw an arc over a. 
Then draw a line through d and c, producing it to meet the last arc in e. A line from 
a to e will be the perpendicular. 

Fig. 12. To divide a straight line a b into any number of equal parts, say four. 
Draw a c at any angle with a b. Lay off on a c four equal spaces. Join b to the Ikst 
point on a c and draw three other lines parallel to this line. They will cut a b into 
the required number of equal spaces. 

Fig. 13. To bisect any angle as a b c. With b as a centre and any radius, draw 
arc a g, then with a and c as centres and any radius greater than one-half a c, draw 
arcs intersecting in d. Draw b d. 

Fig. 14. To construct an angle equal to a given angle e b c. Draw any line as 
d e. With b as a centre and any radius, cut the sides of the given angle in c and e. 
With d as a centre and same radius, draw arc e b. With c e as a radius and e as a 
centre, cut arc e b in b. Join d and b. 

Fig. 15. When two straight lines, as a b c d, cross each other, the opposite 
, angles are equal. 

Fig. 16. To divide a circle a b into twelve equal parts with the T square, and 
30° X 60° triangle. Draw a b with the T square. Next draw e e with the short 
side of the triangle on the T square, and the next longest side directed to the centre 
of the circle. Draw c c and g g with either of the long sides of the triangle on the 
T square, and the other directed to the centre of the circle. Draw d d and f f with 
the shortest side of the triangle on the T square, and the longest side directed to the 
centre of the circle. 

Fig. 17. To draw this square draw the circle first, then with the T square 
and triangle draw horizontal and vertical lines tangent to the circle. 

Fig. 18. To draw a hexagon with the T square and triangle. Draw the circle 
first, then with 30° x 60° triangle and T square, draw lines tangent to the circle. 

Fig. 19. To draw an octagon with T square and triangle. Draw the circle first, 
then with 45° triangle and T square, draw lines tangent to the circle. 

Fig. 20. To draw an approximate ellipse. Let a b be the major axis and 
c d the minor axis of an ellipse. With a^ as a centre and radius a^ d, draw arc d e. 
Join d b. Make d f equal to b e. Bisect b f according to fig. 8, plate 2. Then g b 
will be the radius for the ends of the ellipse, and h d the radius for the sides. 

10 



Fig. 21. To draw an ellipse. Let a b be the major axis and d c the minor axis. 
With d as a centre and radius g a, draw arcs cutting a b in e and f which are the foci. 
Between g and f affld g and e take any point as h, and with f and e as centres and 
radius h b, draw arcs as at a'' b" d'' c'', then with the same centres and h a as radius, 
draw arcs intersecting those first drawn. Take any other points between f and g 
jfflcl g and e, and repeat the operation as is shown. Througli the intersections of 
these arcs draw the curve of the ellipse. 

Fig. 22. To find the true length of a line when its projections are at an angle 
with both planes of projection. Let a' b' be the vertical projection of a line at an>' 



Plato. ^ 




convenient angle, and a^ b^ the horizontal projection at any angle. Draw the hori- 
zontal line g'' C ; project b' to C and a' to g' . Draw b*" b and a*' a perpendicular to 
a** b^, then make b^ b equal to b' C and a^ a, equal to a'' g\ A line from a to b will 
be the true length of the line. If a line is paralled to one coordinate plane and 
oblique to the other, its projection on the plane with which it is parallel is equal to 
the true length of the line in space. 

Fig. 23 is another method to find the true length of a line. Let a' b\ b*' a'' be 
the projections of a line. Draw the horizontal lines f g and b' C . With b' as a centre 
revolve a'' to c"", then it will be parallel with the horizontal plane. Project c' to b on 
a horizontal line through a^. Then b b^ is the true length, and is its projection on 

the plane with which it is parallel. 

11 



Plate 3. 

Fig. 24. To make a mechanical drawing of a block 4'' high by 3" wide and 
I ^" thick. 

NoTK. — When an object revolves parallel with a plane, its projections on that 
plane are in their true length, but when it makes an angle with the plane, its projec- 
tions on that plane will not be in their true length. Therefore the first view of the 
object must be on the plane with which it is parallel, but as all the views in fig. 24 
are parallel with all the planes, we can draw any of them first. In a case of this 
kind however, it would be well to draw the top or plan first. 

Then let g*" g^ and f"" f be the axes about which the planes of projection revolve. 
Draw the plan to the dimensions given. Project a*" to sC in the front elevation and pro- 
duce it to b', making a'' b'' 4'^ long and project d*" to d"" produced to c''. Connect sC d", c' b", 
thus completing the front elevation. Now project a*" to f, and e'' to d, in T i\ With 
centre g and radius g f, draw the arc f c ; now with the same centre and radius g d, 
draw arc d a. Then c projected to the side elevation and intersected by the 
projection of the top of the block aC" dC" in the front elevation, will locate a" in the side 
elevation, and a"" produced to intersect the projection of a will give e''. The projection 
of the bottom of the block intersected by the projection from c and a, will give b'' g''. 
Connect the points and complete the drawing. 

Pig. 25. This is a drawing of the same block inclined at an angle of 30° with the 
side vertical plane. Here it will be oblique to the horizontal and side vertical planes, 
therefore we cannot draw either of these views first. But as it revolves through this 
angle it remains parallel with the front vertical plane, therefore it is on this plane we 
must draw the first view. Having found which plane to draw first, proceed to draw 
the block in full as shown in the front elevation. Then lay off the thickness in the 
plan and side views to the dimensions given, and from the corners in the front view 
project to the plan and side views, as a' to a*", and b'' and a' to a' and b''. Follow 
the instructions of fig. 24 and complete the drawing. 

Figs. 26 and 27 represent the same block in different positions and are drawn 
by the principles illustrated in figs. 24 and 25. 



12 



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Plate 4. 

Figs. 28 and 29 are drawings of a block to be drawn as shown from the instruc- 
tions on figs. 24 and 25, plate 3. 

Fig. 30. This is the same block revolved so as to make an angle of 15° with 
the side vertical plane. But a" b" in this revolution will remain the same distance 
from the front vertical plane, therefore it follows from what has been said respecting 
fig. 25 that the front elevation must be drawn first. Make a'' b'', fig. 30, equal to a'' b'', 
fig. 29, at an angle as shown. Then make sC f^, fig. 30, perpendicular to a'' b", fig. 
30, and equal to a"^ f', fig. 29. The distance a'' g'', etc., is equal to a" g'', etc., fig. 29. 
From these points draw lines parallel to a'' b'', also from b" parallel to a" f''. To 
draw the plan : From a' , g", etc. , project fine lines which will pass through c^, P, a*', g*". 
Now take the distances from the axis g^ g^, fig. 29, to c^, f', a'', g*" and lay off these dis- 
tances from the axis g^' g'\ fig. 30, on the lines already projected. They will locate 
the corresponding letters in fig. 30, and the points being connected will give- the top of 
the block. The bottom is found in the same way. The side elevation can be p'ro- 
jected from the views already drawn if c*" be projected to the axis P f\ and from the 
intersection of P' f'' with g'' g" as a centre, we revolve c^ to g*" g^. Project c^' down 
until it intersects c", projected from the front elevation, which will locate c^ in the 
side elevation. All other points are located in the same way. 

Fig. 31. In this figure the block makes an angle of 15° with the front vertical 
plane. From what has already been said we see that the side view must be drawn 
first. Draw C e" at an angle of 15° and equal to c'' e'', fig. 30. Then draw a perpen- 
dicular c d and also a perpendicular a b from c'' e"", fig. 30, about midway between 
C e"". Make the distance from c'' e" to a" b'', etc., measured on c d equal the corre- 
sponding distances in fig. 30. The distances from c d to C a"", etc., are the same as 
from a b to C a'', etc., fig. 30. Project c'' a"", etc., to the front view and make the 
distance from f' P to C a'', etc., in the front view, fig. 31, the same as from P P to c"" a" 
in the front view, fig. 30. The bottom is found the same way. The top view can be 
projected from the front and side views as the side view was projected in fig. 30. 



14 



Plate 




Plate 5. 

Fig. 32. This is a drawing and development of a vertical cylinder, cut bj^ a 
cutting plane at an angle of 45° as shown. The only explanation required is in- 
struction for drawing the oblique view and the development. To draw the oblique 
view, draw parallel with a'' b" the trace a^ b of the central vertical plane a^ b''. Now 
the points a' b' can be projected to the trace a'' b perpendicular to a^ b". The dis- 
tances on each side of a'' b will be the same as the distances of the vertical planes 
from a" b''. Now pass a number of auxiliary traces in the plan as shown, and where 
they intersect the circle of the cylinder project them down to the base in the front 
elevation and draw traces as shown. Then project their intersection with a" b" to the 
oblique and side views. I^ay off from a'' b and e'' d" the distances from a^' b" to where 
the same traces intersect the circle of the cylinder. Through the points thus found 
draw traces in the oblique and side views, and where they intersect the traces which 
were projected from the front elevation will be points to draw the ellipse of the top 
end in the oblique view, also for the side elevation, which latter will be a circle. 
The ellipse in the oblique view is the true size of the section made by the cutting 
plane. The ellipse of the base in the oblique view is found in the same manner. 
To draw the development, which is the surface unfolded into one plane, draw a hori- 
zontal line in length equal to the circumference of the cylinder by calculation. Then 
divide it into as many parts as the circle is divided in the plan. From each one of 
these points erect perpendiculars the height of which will be equal to the correspond- 
ing lines in the front elevation. Draw the curve through the points thus found. 

Fig. 33. This should be drawn without any further explanation. The 
corners of the prism will answer without any additional traces. 



16 



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¥ig.33 





Plate 6. 

Pig. 34. This is a drawing of a vertical cone of the dimensions given, drawn 
to locate the poitils c and b. Draw the plan and front views first. Now draw a""^ b'' 
which will be a trace of the central vertical plane c^ b'', parallel with the side of the 
cone, and will also be on a plane parallel with the side of the cone. Next draw the 
ellipvSe as instructed in fig. 32, plate 5. Then project the apex to a""^ ; from a""^ draw 
a trace through the apex of both cones and produce it until it intersects the base 
produced, as at a"". Project a' to a'\ From a" draw a trace tangent to the base of 
the cone as at c^. Project c^ to C. Then C projected to the ellipse in the oblique 
view will give c and b, and these are the points to which the sides of the cone a'^V c, 
a''^ b are to be drawn. 

Fig. 35. This is a drawing of a hexagonal prism to the dimensions given. 
From what has already been said, the pupil should not have any trouble in being 
able to tell which view to draw first, and should also be able to finish the drawing 
without further instructions. 

Pig. 36. This is a drawing of a vertical pyramid with a pentagonal base, cut 
by a cutting plane as shown. Draw the base in the plan as in fig. 36 ; then from 
each corner of the base draw lines to the apex h, which will be the corners of the 
pyramid. Draw a horizontal line e'' d^ which will be the base in the front elevation. 
If all the corners of the base in the plan are projected to the base in the front eleva- 
tion as e^ to e", the corners will be located in the front elevation, and lines from these 
points to the apex a'' will be the corners of the pyramid in that elevation. Now pass 
the cutting plane, and if the points where this trace intersects the corners of the 
pyramid be projected back to the corresponding lines in the plan as C to c'' all the 
points can be located in the plan except f '' , which can be taken from the side eleva- 
tion after it is drawn. Next draw a" a' in the side elevation, which will be a trace of 
the central vertical plane g^ g^ . The distance that b'' is from the central vertical plane 
a.^ a" is the same as the distance that b^ is from g^ g*' . All other points are found in 
the same way. From these points, draw lines to the apex. From the point where the 
cutting plane intersects the corners of the pyramid in the front elevation project 
to the corresponding line in the side elevation as c'' from the front to c"" in the side 
elevation. Now if the distance from a'' a^ to f^'be laid off from g^g\ the point f^ 
will be located. Connect all the points and finish the drawing. 

Pig. 37. This is a drawing of a vertical pyramid {fibM^SLU oblique view and 
should be drawn without further instructions. 



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Plate 7. 

Pig. 38. This is a drawing and development of a vertical pyramid with a 
hexagonal base, cut by a cutting plane a7 b^ at an angle of 30° with the horizontal 
plane. In this figure all the points in the front elevation made by the cutting plane 
a" b'' can be projected directly to the plan by projecting each point to the correspond- 
ing line, and similar projection may be made to the side elevation. If the pupil will 
make an oblique view of the top, parallel with the cutting plane a' b'', the true size 
of the section will be obtained. To make the development take the length of the 
corner e'' f " as a radius, and with centre a in the development draw an arc. Lay off 
three times on either side of the central line the length of one side of -the base taken 
from the plan and connect these points with straight lines. Then from these same 
points, draw lines to the apex a. As the line e' f " is the true length of the corners 
of the pyramid, and is the radius used for the development, all other measurements 
must be taken on the same line. With the T square project C & a' to the line e' f. 
Then the distance from the apex C to each one of these points, on the line e' f '', if 
laid off from a, on the development, will give the points from which to complete the 
figure. 

Fig. 39. This is a drawing and development of a vertical cone cut by a cutting 
plane at an angle of 30° with the horizontal plane. All that is necessary here is to 
pass a number of auxiliary planes, and treat them exactly as the lines forming the 
corners of the pyramid, fig. 38. The length of the arc a' b' is equal to the circum- 
ference of the base of the cone, found by calculation. Follow the instruction in 
fig. 38 and complete the drawing. 



20 



Plate. 7 ;^ 




Plate 8. 

Fig. 40. This is a drawing and development of a vertical cone of the dimen- 
sions given, cut by a plane parallel with the base. From what has already been said 
the pupil should be able to draw the three views. The development is made by 
taking the side of the cone 37 b"" as a radius, and drawing the arc a b. The length 
of the arc is to be made equal to the circumference of the base of the cone. Connect 
a and b with a' , and with radius a^ c^' from the same centre as before, draw arc e' d'', 
thus completing the development. 

Fig. 4 1 . This is a drawing and development of a vertical cone of the dimen- 
sions given, cut by a vertical plane }4'' in front of the axis. The section in the front 
elevation made by this plane will be bounded by a curve, and in order to find this 
curve we must pass auxiliary planes. In this case they will be horizontal as shown 
by the traces in the front elevation, but will be circles in the plan. To find these 
traces in the plan project b"" to b^, and with centre g" and radius g'^b*' draw the trace 
a'^b^. Where that trace intersects the line c'^d'' will be points to project back to the 
trace a" b'' and will give c' d', which are points in the curve. All other points are to 
be found in the same way. Through these points draw the curve, which will be a 
hyperbola. 

To draw the development, draw traces as shown in the plan view, which will be 
the traces of vertical planes where they intersect the circle of the base. Project down 
to the base in the front elevation, and from these points draw traces to the apex which 
will be traces of the same vertical planes. Where these traces intersect the curve in 
the front elevation, project the intersection to the outside of the cone, which is the 
line g' h' . Draw the development as directed in fig. 40. On each side of the centre 
line in the development, lay off the arcs h' f', etc., equal to the arcs h^f^, etc., and 
from these points draw lines to the apex, which will be traces of the same vertical 
planes. The points where the curve intersects these same traces in the front eleva- 
tion are already projected to the outside of the cone g' h'\ From g^ to each one of 
these points in g'' h'' will be the distance to lay off from the apex on the corresponding 
trace in the development. Through these points draw the curve of the development 
and thus complete the drawing. 

Figs. 42 and 43 are drawn in the same way. 



22 



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. Plate 9. 

Fig. 44. This is a drawing of a vertical cone to the dimensions given, and is 
drawn the same as the figures on plate 8. The oblique view is on a plane parallel to 
the cutting plane in the front elevation. 

Pig. 45. This is a drawing illustrating the same principles as fig. 41. 

Fig. 46. This is a drawing of a vertical cone intersected by a horizontal cylin- 
der. To do this draw all three views in full first. Then in the side elevation draw 
traces a' b' , a' d' tangent to the cylinder. The distance from the central vertical plane 
to b"" laid off perpendicular to the central vertical plane in the plan will give the 
point b** in the plan. Draw trace b^'a*'. Project b'^to b' in the front elevation and 
draw the trace a'' b'. Then C in the side elevation projected to the corresponding 
trace a' b' in the front elevation will give the point C, which is one point in the curve, 
and projected to the same trace in the plan will give c^ . All other points are found 
in the same way by projecting the intersection of the traces with the cylinder in the 
side elevation to the corresponding traces in the other views. 

Fig. 47. This is a drawing of a vertical pyramid with a square base, inter- 
sected by a horizontal cylinder as shown. Draw the trace a' b' in the side elevation 
tangent to the cylinder. Then a' b' in the front elevation will be a trace of the same 
vertical plane. Draw as many more traces as are needed to find the curve of inter- 
section, and proceed as in fig. 46. 



24 



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Plate 10. 

Fig 48. This is a drawing and development of a vertical cylinder intersected 
on the right by a horizontal cylinder, and on the left by another horizontal cylinder 
as shown. If we draw a circle to the right in the plan to represent the end of the 
horizontal cylinder, and divide that circle into, say twelve equal parts, and through 
these points draw horizontal lines and produce them to the left until they intersect the 
vertical cylinder as shown, these lines will be traces of vertical planes, and a^ b*" will 
be at b'' c"", a' d' in the front elevation. Where the same trace intersects the vertical 
cylinder in the point c" , project c^to the front elevation, which will intersect b' c' in 
0"" and a' d' in d'', then C d' are points which are common to both surfaces, and, 
therefore, points of intersection of the cylinders. All other points for both cylinders 
are to be found in the same way. 

To draw the development, make the line e'' e"" in length equal to the circum- 
ference of the vertical cylinder, and e' e" equal to the height. Connect e'' e"", e'' e'' , etc. 
Make e' g'' equal to one-fourth of the circumference of the vertical cylinder, by 
making e'' h' equal to f'c'' and h'' c', etc., equal to c'^g'' , etc. From the points thus 
found erect perpendiculars as shown, and as many on the left of g'', as there are on 
the right. The distance g' a'' is equal to g'' C. All other points in the curve are 
found in the same way. Make g" f ' equal to one-half of the circumference of the 
vertical cylinder. If the pupil understands what has already been said, there should 
be no difficulty in finishing this and drawing the other developments of the horizon- 
tal cylinders. 

Fig. 49. This is a drawing and development of a vertical cylinder intersected 
on the right by a cylinder oblique to the horizontal plane, and parallel with the verti- 
cal plane of projection. It can be drawn by a close study of fig. 48. 



26 



Plate. 10 




Fig-.f^ 






Plate 11. 

Fig. 50. This is a drawing and development of a horizontal cylinder intersec- 
ted by two vertical cylinders. The axis of the 2 inch vertical cylinder is ^ inch 
in front of the axis of the horizontal cylinder, and lyz inches from the end. The 
axis of the 1)4. inch cylinder is ^ inch behind the axis of the horizontal cylinder 
and I % inches from the end. By attentive study of this plate in connection with 
the explanation of plate 10, there should be no difficulty in making this drawing and 
the developments. 



Plate 12. 

Fig. 5 1 . This is a drawing and development of a vertical cylinder cut by a 
cutting plane making an angle of 6o° with the central vertical plane, and of dimen- 
sions and in positions shown. Divide the distance between the traces at the top and 
bottom of the cylinder in the plan into, say two equal parts, and draw the trace as 
shown. Divide the height into the same number of equal parts and draw the trace 
as shown. Make the projections to the front and side elevations, and through the 
points thus found draw the curves. To draw the development : — Draw traces in the 
plan of a number of vertical planes, say twelve, and project them to the front eleva- 
tion. These traces are not shown, but by referring to the development, and turning 
back to fig. 32, plate 5, the pupil should have no difficulty in finishing the drawing. 

Fig. 52. This is a drawing and development of an inclined cylinder with the 
base and top parallel with the horizontal plane, the axis making an angle of 60° 
with the horizontal plane and parallel with the front vertical. This will be left to the 
pupil to finish. 



30 



P l at e . /^ 



Figs/ 




FigS^ 




' 7 ' I 





Plate 13. 

Fig. 53. This is a drawing of a vertical hexagonal pyramid intersected by a 
horizontal cylinder. Draw all three views in full first. Divide each side of the base 
into four equal parts. From these points draw traces to the apex, which will be 
traces of vertical planes. Project their intersection with the base, to the base in the 
front elevation. From these points in the base draw traces to the apex, which will 
be traces of the same vertical planes. The traces in the side elevation will be the 
same distance from the central vertical plane a' b'' as they are from the central vertical 
plane a^ b^in the plan. Where these traces in the side elevation intersect the cylinder, 
project that intersection to the corresponding trace in the front elevation and plan as 
c' c'', c' c^,and complete the drawing. 



32 



I 



Plata. /3 




Plate 14. 

Fig. 54. This is a drawing and development of a vertical hexagonal pyramid 
intersected by a cylinder. The axis of the cylinder is parallel with both planes of 
projection and % inch in front of the axis of the pyramid. Draw all three views in 
full, and proceed as instructed in fig. 53, plate 13, except the trace sC b', which is 
tangent to the cylinder in the side elevation, a^is the same distance from the central 
vertical plane as a" is from f "". The point C in the side elevation can be found by 
drawing a line through the centre of the cylinder and perpendicular to b^ a"". Then 
if it is projected to the same trace in the front elevation it will give C, and projected 
to the plan will give c^ . To draw the development of the pyramid, with b'' h'' as a 
radius and centre b" in the development, draw an arc. With a distance equal to one 
side in the plan, lay oiF six spaces, connect these with straight lines, and also draw 
lines to the apex b"", now these lines are equal in length to b"" h'' in the side elevation, 
and are also the corners of the pyramid. If the points where the lines which repre- 
sent the corners of the pyramid intersect the cylinder be projected to b"" h", the dis- 
tance from b' to these points will be the distance to lay off on the corresponding lines 
in the development from b"". On account of the curve in the development it will be 
necessary to draw lines in the development midway between those already drawn, 
which will be the same as the traces in the plan midway between the corners. Now 
b'' h" is the true length of the corners, because it is parallel with the plane on which 
it is projected, therefore b"" e^ is the true length of b^ e^ and all other lines midway 
between the corners. If the points where the curve of intersection intersects these 
traces in the front elevation be projected to b'' e'', the distance from b'' to these points 
will be the distance to lay off on the corresponding line in the development from b"". 
The line g'^c^is the distance in the development that C is from b", and is found by 
making g^b^ equal to C f"" and perpendicular to b'^a'*. Connect g'^c'', draw curves 
through the points in the development and complete the drawing. 



34 



Plat.fl it 




Plate 15. 

Fig. 55. This is a drawing of a vertical pyramid with a triangular base, and 
the front side making an angle of 15° with the front vertical plane, intersected by a 
triangular prism with its axis parallel with both planes of projection, the front side 
making an angle of 15° with the front vertical plane and drawn to the dimensions 
given. If we use the line f " g", which is the top corner of the prism, as the trace of a 
cutting plane, it will give a triangular section in the plan as a^ b*" c*" . Where the 
traces of this section intersect the line f^g^ (which is the same corner of the prism) 
will be points of intersection of the two solids, because f '^ g^ and a*" b^ c^ are in the 
same plane, and these points d** e'' projected back to f "^ g", will give the points d" e"" in 
the front elevation. All other points are found in the same way, except where the 
left corner of the pyramid is cut. Use the line forming the corner of the pyramid as the 
trace of a cutting plane, and where it intersects the lines of the prism project to the 
corresponding lines in the plan, which will give a section of the prism in the same 
plane as the corner of the pyramid is. Where the traces of this section intersect the 
corner of the pyramid will be the required points of intersection, and if projected 
back to the front elevation, will give the points there. The pupil should study these 
principles, as he will have many cases of their application. 



36 



Figf.iT 



Pla te, 1^ 





Plate 16. 

Pig. 56. This is a drawing of a vertical pyramid with triangular base, making 
an angle of 15° with the side vertical plane, intersected by a triangular prism, the 
axis of which is parallel with the vertical plane, ^ inch in the rear of the axis of the 
pyramid and making an angle of 30° with the horizontal plane. The bottom side 
makes an angle of 15° with the horizontal plane. The intersection of these two 
solids can be found in the same way as fig. 55, plate 15, but there are other ways, 
and it would be well if the pupil would use other lines as traces of cutting planes. 
Suppose we use the line a^ b"" as the trace of a cutting plane. Where that trace 
intersects the base of the pyramid in the plan it should be projected to the base in 
the front elevation, and where the same trace intersects the front corner in the plan it 
should be projected to the corresponding line in the front elevation. Draw traces 
from that point to the points already found in the base, and where these traces inter- 
sect a' b^ will be the required points of intersection, because they are in the same 
plane. If the pupil will have patience to find one point at a time in making these 
drawings, any of them should be handled without difiiculty. 



38 



Pl a t o . /6 



Figr.5'6 



-;^ 




T 






_>L 



Plate 17. 

Fig. 57. This is a drawing of axes or centre lines oblique to both planes of 
projection, about which any solid may be drawn very readily. Draw b'^ b', which 
will be the trace of a horizontal plane ; then f f perpendicular to b"" b"" and a"" c' at the 
desired angle (45°) with the horizontal plane, making it of any convenient length longer 
than the length of the object to be drawn. Draw also c'' c'^any convenient length. 
Now draw c'^a, making the desired angle with the front vertical plane. From a*" to 
h'' draw a line perpendicular to c*" a and g'' e'' parallel to c*" a, and any convenient dis- 
tance from c^ a. Then c^ e'' must be parallel to a*" h''. Make g"" h'' equal to the verti- 
cal distance that c^ is from b"" b"". Draw d' d^ through h'' and parallel to g"" e^, which 
will.be a trace of the horizontal plane b"" b''. Now draw a line from h" to e", which 
will be the true length of the axis a" c'', and will also be a parallel view of the axis. 
Here we must draw the object first in its true size. The line c" f '' may, for conven- 
ience in drawing, be equal to C c'' . If c'' c'' be produced to intersect a horizontal from 
a'' as at b, then the vertical distance C" b, laid off on c"" f'' produced as at c and pro- 
jected to b' b"", will give f "^j and f "^ f "^ is a side view of the axis. 

Fig. 58. This is a drawing of a solid inclined to both planes of projection as 
shown. The axis must be drawn in full first, as instructed in fig. 57. If it is true 
that the distance g"" h' is equal to the vertical distance from C to b'' b"", it is also true 
that the vertical distance from C to any point in the front or side views is equal to 
the distance from g^ e"" to the corresponding point in the auxiliary view, taken per- 
pendiculi)^^ to g' e'. Draw the auxiliary first as shown. The top view can be pro- 
jected from the auxiliary view, and the thickness taken from the end view. The 
front view can be projected from the top, and the vertical distances of all the points 
found as already explained. The side view is not shown, but must be drawn by the 
pupil. It can be done by projecting all the corners from the top and front views. 

Fig. 59. This is a drawing of a triangular prism of the dimensions shown, and 
can be drawn from instructions already given. 

Fig. 60. This is a drawing of a cube of the dimensions given, and if the 
others on the plate are understood there will be no difficulty in drawing it. A 
thorough knowledge of these principles is something that no draughtsman can afford 
to be without ; therefore it is hoped that the pupil will give this plate undivided atten- 
tion and study. 



40 






Fig-.J/ 



^ 



P lat e . /^ 






h^ef 



--t-h^ 



mgsf 




Fig-.i^^ 




Fig". 60 




Plate 18. 

Fig. 6 1 . This is a drawing of a vertical cone intersected by a vertical cylinder, 
the axis of which is % inch behind the central vertical plane. 

Pig. 62. This is a drawing of a vertical cone intersected by a horizontal hexa- 
gonal prism, the axis of the prism piercing the axis of the vertical cone i^ inches 
above its base. 

Pig. 63. This is a drawing of a vertical hexagonal prism intersected by a hori- 
zontal hexagonal prism as shown. 

Pig. 64. This is a drawing of a vertical prism intersected by a horizontal 
prism as shown. 



42 



Figr.6/ 




:Plate- , f^ 



Figr.65^ 




-J 



Figr.63 /-^ 




Fig". 6'/ 




Plate 19. 

Pig. 65. This is a drawing of a vertical hexagonal pyramid intersected by a 
horizontal prism 2^ inches square, the axis of which is i inch behind the axis of 
the pyramid. 



44 



Fiir.6^ 



Plate. /f 



_ J _ - 



'>--f--;. 





e -> 




Plate 20. 

Fig. GQ. If a circle is rolled on the outside of another circle, a point on its 
circumference will trace a curve which is called an epicycloid. If a circle is rolled on 
the inside of another circle, a point on its circumference will trace a curve which is 
called a hypocycloid. To describe these curves, draw the large circle as shown and 
a horizontal and vertical centre line. Then, setting the spacing dividers to any dis- 
tance, say j\ of an inch, step this distance on the large circle, starting at the inter- 
section of the vertical centre line with the circumference, and step both ways. Be care- 
ful not to change the distance of the dividers. From the centre of the large circle, 
draw the paths of the centre of the small circles as a b c d. From the points laid off 
on the large circle draw radial lines as shown, and from where they intersect the path 
of the centres of the small circles as centres, and with radius equal to the radius of 
the small circles, draw arcs as shown. Now with the dividers at the same distance, 
step back on each arc from the points on the large circle as many times as the point 
numbers from the vertical centre line. Then through the last points thus found draw 
the curve. 

If a cord is kept taut while it is unwound from a cylinder, a point at its end will 
trace a curve which is called an involute. To describe an involute : — With the spac- 
ing dividers set at about -^ of an inch, step this distance on the large circle as shown. 
Be careful not to change the distance of the dividers. From these points draw radial 
lines and a perpendicular to each of these lines tangent to the points on the large circle. 
With the dividers at the same distance, step back on each tangent as many times as 
it is from the vertical centre line. Through the last points thus found on each 
tangent, draw the curve. 

Fig. 67. If a circle is rolled on a straight line, a point on its circumference will 
trace a curve which is called a cycloid. To describe the curve draw the straight line 
h h and lay off spaces the same as in fig. 66. Through these points draw perpen- 
diculars. A straight line through the centre of the circle will be the path of the cen- 
tre. The intersection of this path with the perpendiculars will be centres from which 
to draw arcs with a radius equal to the radius of the circle. With the dividers at 
the same distance, step back on these arcs as many times as the arc is from a"" a''. 
Through the last points on each arc thus found draw the curve. 

These circles which we have been rolling are called generating circles, and are 
used to generate the curves of gear teeth. In good practice they are 1.9098 times the 
circular pitch in diameter, which is the length of an arc of the pitch circle from the 
centre of one tooth to the centre of the next tooth. 



46 



PlRtft -^0 



Fig-. 6 6 




Plate 21. 
CONVENTIONAL SCREW THREADS. 

The bolt at the top of this plate is a drawing of a U. S. Standard bolt and nut, 
with the conventional thread. The sides of the thread known as the U. S. S. make 
an angle of 60° with each other, with a fiat at the top and bottom of the thread equal 
to ^8 of the pitch, and have a depth equal to .65 of the pitch. The pitch of a screw 
is that distance which it travels through the nut in one revolution. To lay out and 
draw this thread : First, draw the bolt in full (referring to table No. 5 for all 
dimensions not shown on the drawing) and lay off the depth of the thread arid bisect 
it and draw what may be termed a pitch line as shown in the enlarged thread below 
the bolt ; also draw a line at the root or bottom of the thread. Now it remains to 
determine the pitch, which is as follows : This bolt has 3)^ threads per inch, and 3^ 
threads are equal to 7 threads in 2 inches, which is equal to j pitch. The pitch being 
determined, lay off 2" on the lower pitch line and divide into 14 equal parts, then 
each division will be equal to one-half of the pitch. L/ay off as many more of these 
divisions as may be required for the number of threads on the bolt, and project each 
one of these points to the pitch line at the top of the bolt. Now with the 30°x6o° 
triangle with the shortest side to the T square, draw lines through alternate points on 
the lower pitch line, then in the opposite direction through the remaining points. 
The distance between the lines where they intersect the outside and root lines will be 
the width of the flat at top and bottom, which will be }i of the pitch. Draw the top 
side in like manner, care being taken to have the outside of the thread directly 
opposite the root at the bottom side of the bolt, as shown. Having completed this, 
connect all points at top and bottom with straight lines as shown in the plate. 

The bolt at the bottom of the plate represents the conventional square thread, 
the depth of which is equal to one-half of the pitch. After the outlines of the bolt 
have been drawn in full, lay off on the bottom side of the bolt /\!' and divide into 14 
equal parts, because 1 3^ threads per inch are equal to 7 threads in 4'' or ^ pitch. 
Then each division will be equal to one half of the pitch. Lay off as many more 
divisions as are required for the number of threads on the bolt, and also from the top 
and bottom side lay off the depth of the thread equal to one division, and through 
these points draw lines which will represent the root diameter, also lines through the 
points in the bottom side of the bolt, from the outside to the root both top and bottom. 
These lines will represent sides of the threads on the central plane and must be 
perpendicular to the axis of the bolt. Then connect each point in the top and bottom 
sides of the bolt with straight lines, the first one at an inclination toward the left 
equal to one division, and draw all others parallel to this one. They represent the 
outside of the threads. Then at the same inclination in the opposite direction, draw 
the short lines on the left of the thread at the top, and on the right of the thread at 
the bottom, which will be an outline of a portion of the thread on the underside of 
the bolt. Now from where the lines forming the sides of the thread on the central 
plane intersect the lines forming the root diameter, draw lines to represent the root 
of the thread as shown. These are only drawn to the axis, because from this point 
on they could not be seen, and if drawn would be dotted, which is not desirable. 

48 



Plate. 2] 




k-piich->| 




/ 

\ 




'i 


"T"I1 


\ 




111 


< S:f > 


M 




i 


U 



1^ threads pen inch. 




- -% ^ 



Plate 22. 
THE HELIX. 

If we suppose that the line a b is parallel to the axis of the cylinder, and a point 
moving along the line from a to b at a uniform velocity, while the cylinder makes 
one revolution about its axis at a uniform velocity, the point, if touching the cylinder, 
will trace a curve known as a helix. The distance a b will be what is termed the 
pitch of the helix. The pitch of a screw is equal to the axial distance through which 
the generating point travels in one revolution of the cylinder, both moving at uniform 
velocities. The helix may be drawn as follows : Draw a cylinder 4 inches in 
diameter, and 4 inches long, and divide the circumference of the cylinder in the top 
view into 12 equal parts, subdividing 4 of these into 3 equal parts as shown. Also 
divide the length of the cylinder into the same number of equal parts, and subdivide 
as is clearly shown in the plate. This may be done by drawing the straight line 
b h perpendicular to the axis of the cylinder, and in length equal to the circumference, 
laying off the distances in the top view on b h, and from each point drawing perpen- 
diculars as shown. Also draw h a, which will be the development of the helix. 
Then by projecting the intersection of the perpendiculars with h a, as is shown in 
the plate, the length of the cylinder will be properly divided. These traces intersected 
by projections from the top view, will locate the points through which to draw the 
helix a d b. The inner circle in the top view represents what is termed the * ' root 
cylinder " of a screw thread, the helix of which is drawn in the same manner as the 
above. From the projection in connection with the numbers there should be no diffi- 
culty in making this drawing. The angles shown in the plate are the angles which 
a tangent to the helices makes with the plane of the base of the pitch cylinders. The 
co-tangent of the angle is equal to the circumference of the cylinder divided by the 
pitch. The drawing of a quadruple square thread screw to the right of the top view 
gives the application of the helix to the screw thread. 

This being a quadruple screw, means that the pitch is divided into four equal 
parts, each part being equal to one thread and one space, and is termed " individual 
pitch," and as the pitch of the helix is 4 inches, the individual pitch will be i inch. 
Make the thickness of the thread, the width of the space and the depth, each equal 
to one-half of the individual pitch. 

Then from a piece of thin wood make a curve to fit the helices from a to d and 
from c to e, and with these draw the helices of the screw as clearly shown. This 
being a quadruple screw a thread at the top will be opposite a thread at the bottom. 

I^ay out and draw^ the nut with the same curves. 

In drawing the helices ol a screw it is well to use a bow pen to draw a small 
portion at the point of tangency, as at g. 

The drawing at the lower side of the plate is to illustrate the method of drawing 
the thread usually used for a worm to work with a worm wheel. The helices for 
this thread are determined in the same manner as explained above. To draw this 
thread draw the outside, the pitch and root cylinders as shown at the right, and the 
outside, the pitch and root diameters in the front view. Then lay off on the pitch 
lines the pitch and thickness of the teeth equal to one-half of the pitch, and through 
these points draw the sides of all of the teeth as shown in the section and the dotted 

50 



I 



outline at the bottom side. From the points where the sides of the teeth intersect 
the outside, the pitch and root cylinders, draw the outside, the pitch and root helices, 
then straight lines tangent to these helices, will be the visible contour of the thread 
in the finished drawing, instead of the straight lines drawn first through the pitch 
points in the central plane. It will be noticed that these tangents at the right of the 
thread at the top extend to the point of tangency as at v v, and only to the root 
on the left ; at the bottom this will be reversed, as at h h. 



Plate. 2 2 



SCREW THREADS. 



^ s 




After the tool to cut this thread has been filed to the proper angle, file off the 
end of the tool until a gauge, as at a — which has a width equal to one-half of the 
pitch and a depth equal to .35 of the pitch — will touch the end and sides of the tool 
at the same time. It will then be the correct shape to form the thread, if the top side 
of the tool is in the same plane as the axis of the screw. When the thread is finished 
a gauge as at b should touch the top and sides of the thread at the same time. This 
gauge has a width equal to one-half of the pitch, with a depth equal to .3 of the pitch. 

The dotted outlines are not to be drawn, as they are only shown in the plate 
for illustration. 



51 



Plate 23. 
SPUR GEARING. 

When motion is transmitted from one shaft to another, the axes of which are 
parallel, toothed wheels are used, the pitch surfaces of which are imaginary cylinders." 
lyCt two perfect cylinders be keyed fast on shafts whose axes are parallel and such a 
distance apart that the cylinders are in contact. Then if one cylinder revolves the 
other will revolve also, and if there is no slipping the two cylinders will have equal 
velocities at the point of contact. On these cylinders is laid off the pitch, which 
is the distance from the centre of one tooth to the centre of the next, measured along 
the pitch circle. It is termed circular pitch. The addendum is generally .3 of the 
circular pitch, and .the space below .35 of the circular pitch, making a total depth of 
.65 of the circular pitch. The diameter of the pitch circle is equal to the number of 
teeth multiplied by the pitch, and this product by .3183. Then pitch diameter = 
number of teeth X pitch X -3183. Outside diameter equals pitch diameter plus .6 of 
the pitch, and the root diameter equals pitch diameter minus .7 of the pitch. The 
radii of these circles equal the diameters, divided by 2. 

Diametral Pitch. — If a wheel has 48 teeth and 6 inches pitch diameter it is 8 

pitch. That is, for each inch of the pitch diameter there are 8 teeth on the circum- 

r ATA. T^-^ 1, number of teeth 

ference. Then Pitch = — ^7-:; — -r- 1 — and 

pitch diameter 

•r^., , .. , number of teeth 

Pitch diameter = t-—. 

pitch 

Number of teeth = pitch diameter X pitch 

I 



The addendum is 
Then whole diameter 
Whole depth of the teeth 



diametral pitch 
number of teeth + 2 
diametral pitch 

2-15708 
diametral pitch 



Pitch = 8. Number of teeth := 48, then ^ = 6" pitch diameter. 

8 

And ^ = 8 pitch. 

And outside diameter == "^ — - — 

Involute Gearing. — Compute the diameters as explained for circular pitch, and 
lay off on the pitch circles of the wheel and pinion the pitch, also the thickness of the 
teeth equal to one-half of the circular pitch. Then draw through c a straight line 
making an angle of 14^2° with the centre line, and perpendicular to this line draw the 
line of action through the pitch point p. The intersections of these two lines will 
locate the point b. With c as a centre and radius c b draw base circle as shown, 
and divide b p into say five equal parts. Then starting at b lay these same parts off 
on the base circle toward the left. The last point will fall a little to the left of the 
centre line. Also mark off four or five times from b toward the right. I^ay the 
dividers down carefully so that their distance will not be changed. From these points 
draw radial lines and a perpendicular to each of these, tangent to the points on the 

52 



base circle. With the dividers which have not been changed, step back on each 
tangent as many times as it numbers from p. Through the last point on each tangent 
thus found draw the involute which does not extend below the base circle. Find 
•by trial centres from which face and flank curves can be drawn, then through these 
centres draw circles : which will be circles of centres from which all the curves may 
be drawn with the same radius. The flanks below the base circle will be radial lines. 
The radius for the fillet is equal to the product of the constant number for 30 teeth in 
column No. 5, table No. 3, multiplied by the pitch. The involute for the pinion is 
drawn in the same manner. The flanks of the teeth are drawn from the base circle 
tangent to the circle at the centre, the radius of which is equal to the product of the 

Plato. Z3 



INVOLUTE GEARING 




J/_ V ■ ' I- A-i' /. L.-yh i . no ■ 



constant number for 12 teeth, column 4, table No. 3, multiplied by the pitch. Draw 
a few teeth of the pinion in gear with the wheel as shown at the left hand corner at 
the bottom. Also the line of action as shown by the heavy lines. The straight line 
g h represents the flexible cord, which being unwound from the base circles generates 
the involutes. This line is also the path of contact of the teeth, that is the teeth are 
in contact with each other along this straight line. The rack is a wheel with an 
infinite number of teeth ; the involutes of the flanks are straight lines through the 
pitch points, perpendicular to the line of action, that is 14-/4° in opposite 
directions. The faces of the teeth are drawn with a radius equal to the product of 
the pitch and the constant number for 150 teeth, column i, table No. 3, laid off from 
the pitch point p on the line of action produced. 

53 



Plate 24. 
APPROXIMATE INVOLUTE TEETH. 

Compute the pitch, the outside and root diameters of the wheel and pinion as 
explained in plate 23, and draw the line of action at 14)^°, which may be done as 
follows : b c is equal to the cosine of 14/^°, and p b is equal to the sine of 14/^°. 
As .96815 = cosine of 14^° and .25038 =^ sine of i^%°, p c X .96815 = b c, and 
p c X .25038 = p b. This is for 14^^ ° only. For 20° it will be p c X .93969 = b c, 
and p c X .34202 = p b, and this can only be used for 20°. The angle of action being 
drawn, lay off the pitch, and thickness of teeth equal to one-half of the pitch. The 
wheel has 30 teeth and is 2 inches pitch : find in table 3, column 2, opposite 30 teeth, 
the constant number, 1.4079, which, multiplied by the pitch, will give the radius 
with which to draw the face of the teeth. This radius must be laid off on the line of 
action from the pitch point p toward g"' and will give the centre from which to draw 
the face of the teeth, as p v. Through this centre draw a circle which will be a circle 
of centres from which to draw the remainder of the teeth. Find in table 3, column 3, 
opposite 30 teeth, the constant number .990, which, multiplied by the pitch, will give 
the radius for the flanks of the teeth from the pitch circle to the base circle. This 
laid off on the line of action from the pitch point p toward b will give the centre from 
which to draw the flanks. Then a circle through this centre will be a circle of centres 
from which to draw the remainder of the teeth. The teeth of the pinion are drawn in 
the same manner, as the wheel. By referring to the table and taking the constant 
number opposite the number of the teeth desired, a wheel having any number of teeth 
may be drawn, the teeth of which will be as accurate as the average teeth in actual 
practice. The rack teeth are drawn in the same manner as in plate 23. The object 
in rounding the face of the rack teeth is to make them interchangeable with a 1 2 tooth 
pinion. The space between the teeth of a 12 tooth pinion below the base circle is 
almost parallel, being just a trifle wider at the base circle than at the root circle. 
This is done to facilitate matters in cutting the teeth, and also adds some strength 
to the teeth. If the flanks below the base circle were made so that the rack teeth would 
work without rounding the face, (which should not be done), then the space between 
the pinion teeth would be wider at the root than at the base circle, which would com- 
plicate matters in cutting the teeth and also weaken the teeth. Drawing the teeth as 
they are, not only necessitates the rounding of the faces of rack teeth but all wheels of 
150 teeth and over must be drawn with the radius for 150 teeth. By increasing the 
angle of action to 20° or more this is overcome, allowing all the wheel teeth to be true 
involutes, and the sides of the rack teeth straight lines, as they should be, but for the 
reason already explained. 



54 



Plata Z4 




Plate 25. 
CYCLOIDAL GEAR TEETH. 

As already explained in plate 20, if a circle rolls on the outside of a circle, a fixed 
point on its circumference will generate an epicycloid, which being on the outside of 
the circle will be the curve which must be used for the faces of the teeth. If a circle 
rolls on the inside of a circle, a fixed point on its circumference will generate a hypo- 
cycloid, which, being on the inside of the circle, will be the curve which must be used 
for the flanks of the teeth. And if a circle rolls on a straight line, a fixed point on its 
circumference will generate a cycloid, which will be the curve for the face and flank 
of the rack teeth. Compute and draw the pitch, the outside and root circles for a pinion 
having 12 teeth and a wheel having 30 teeth 1% inches pitch, also the pitch the out- 
side and root lines for the racks as shown in the plate. Then draw the generating 
circles tangent to the pitch circles at the pitch points p and p, and, using the pitch point p 
as the fixed point on the generating circles, generate the curves as explained in plate 20. 

The diameters of the generating circles are equal to half the pitch diameter of the 
pinion, which will make them equal to 1.9098 times the pitch. Then the diameter of 
the generating circle for a wheel of any pitch, is equal to 1.9098 X pitch. The curves 
for the rack teeth are generated in the same manner as the cycloid in plate 20. All 
the curves being generated, lay off the pitch, and the thickness of the teeth equal to 
half the pitch, and find centres to approximate these curves, and through these centres 
draw circles, which will be circles of centres from which to draw the curves of the 
teeth. The path of contact is represented by the heavy arcs a p, p b, that is as the 
wheels roll the teeth are in contact along these arcs. By multiplying the proportional 
figures in the plate the different proportions of the wheel may be obtained. Thus 
.625 p means that the thickness of the rim below the root circle is equal to .625 X 
pitch, and the rest in like manner. The width of the arms is equal to 1.625 times 
the pitch at the pitch circle, and enlarged toward the centre .75 in 12 inches. The 
thickness of the arms at the pitch circle is equal to .4 of the pitch, and is enlarged 
toward the centre .75 in 12 inches. A normal section of the arms should be an ellipse. 
The radius of the fillets at the root of the teeth may be equal to .05 of the pitch. 



56 



Plate 26. 
BEVEL GEARING. 

It will be readily understood from what has already been said about spur gearing 
that when motion is transmitted from one shaft to another by such gearing, the shafts 
must be parallel to each other. When shafts are so situated that the axes of the 
shafts are in the same plane and at right angles to each other, they will, if produced, 
intersect each other. Motion is transmitted by wheels whose pitch surfaces are cones, 
instead of cylinders. They are known as bevel gears, if the cones are unequal ; and 
as mitre gears, if the cones are equal. To draw the bevel gear : First draw the axes at 
right angles to each other ; and lay off from the intersection of the axis at o on the 
axis of the pinion the pitch radius of the gear, as o b ; and the pitch radius of the 
pinion on the axis" of the gear, as o h. Now draw lines through b and h until they 
intersect in g and make b b" equal to b g ; also h h" equal to h g. Then g b'' will be 
the pitch diameter of the pinion ; and g h'' the pitch diameter of the gear. From the 
points g b"" and h'' draw lines to the point o. The angle g o h is the centre angle of 
the gear ; and gob the centre angle of the pinion. These angles may be found as 
follows : When the shafts are at right angles to each other only, divide the number 
of teeth in gear by the number of teeth in pinion, which will give the tangent of the 
centre angle of gear. This angle, subtracted from 90°, will give the centre angle of 
pinion. Now through the points g b"" and h'' draw lines perpendicular to go, b"" o 
and h"^^ o, and produce them until they intersect the axis as at o'' and g"". These 
lines will represent the normal cones to the pitch cones o h'' g and o g b''. Lay off 
on these perpendiculars as shown at h^ the addendum and depth below — which is the 
same as in spur gearing — and from these points draw lines to o. Then lay off the 
length of the teeth, 2 inches, and draw lines through these points parallel with the 
lines forming the outside end of the teeth. If these normal cones be developed the 
base of each one will be an arc of a circle. The centre of one is o'' and radius 0"" g ; 
and of the other the centre is g" and radius g'' g. Now produce o g as to h", and from 
o'' and g'' draw lines parallel to o g h^\ Then at any convenient distance draw c c'' 
parallel to o" g'. It is here we develop the outline of the teeth. Draw the line of 
action at an angle of 20°. Now with c as a centre and radius c p draw arc of pitch 
circle ; and with the same centre and radii found in the same manner as c p draw 
addendum and root circles. The pinion is treated in like manner. Lay off the pitch 
and thickness of teeth and proceed exactly the same as for spur gearing, with the 
exception of one thing, which is the number of teeth from which to select the tabular 
number for the faces and flanks of the teeth, h g is the true pitch radius of this gear 
having 18 teeth of i'' pitch ; but the teeth are laid out on an arc of a circle of a much 
larger radius, which is c p. Then a gear of radius c p will contain i'' pitch more 
than 18 times. Therefore we must select a tabular number for a gear of a greater 
number of teeth than 18. And also a cutter to cut the gear for a greater number 
than 18. This is determined as follows : Multiply c p by 6.28 and divide by circular 

pitch, then we have : Number of teeth = -^ — -. -. — ;- 

circular pitch 

Use tabular number in table No. i, 20° line of action, and proceed in the same 

manner as for 1^%° . The thickness for the small end of the teeth is shown on the 

arc V v, the radius of which is equal to r v. This pitch is also equal to the product 

of the pitch at the large end and the quotient of o g into o v. The teeth at the small 

58 



end are drawn in the same manner as at the large end. The same tabular numbers 
must be multiplied by the pitch on the arc v v. 

To draw the view at the left : Project from the section of the pinion as indicated, 
and through these points draw the pitch, the outside and the root circles. Next lay off 
the pitch as many times as there are teeth in the pinion ; also the thickness of the 
teeth, equal to one-half the pitch. Now make projection from the small end of the 
teeth in the sectional view, in the same manner as from the large end, which will 
give points through which the pitch, the outside and the root circles of the small 
end will pass. 

^ Plate, za — -i.— — 



BEVEL OEAHINO 




To draw the outlines of these teeth : Draw one tooth as at g^ from the explanation 
of the enlarged tooth below. Bisect the arc f" f"" with the radial line e'' e'', and make 
the distance each side of e^ e'' on the outside, the pitch and the root circles equal to 
the distance each side of 0"" c''. Through these points draw the outlines a'' f ' d'' ; a^' f '' d''. 
Then where radial lines through these points intersect the circles at the small end of 
the teeth, will be shown the outline as indicated on one side by a f d, and also the 
outside and root of the teeth as sC a and d"" d. Find, by trial, centres from which 
arcs can be drawn through a'' f "" d"" and a f d, then circles drawn through these centres 
will give circles of centres from which the remaining teeth can be drawn through 
the pitch point. All elements of the teeth of bevel gear wheels must run to the 
vertex o, of the pitch cone, whatever may be the angles of the shafts to each other. 
The teeth in the upper half of the pinion are drawn as indicated by the projection. 

59 



Plate 27. 
WORM AND WORM WHEEL. 

This gearing is employed wben the shafts cross each other but do not intersect, 
and may be considered as a spur wheel and rack ; that is, a section on the central 
plane b b is the same as a rack and spur gear having the same pitch and number of 
teeth. Therefore draw the pitch circle of the gear ; and tangent to it at the pitch 
point p draw pitch line which may be considered as the pitch line of a rack to work 
with the gear. From this pitch line lay off the pitch diameter of the worm 3.375''. 
Draw the outside and root diameter of the gear and worm in the manner instructed 
in plates 22 and 23. Next draw the line of action as shown, and generate the involute 
in the manner already explained. Next divide the wheel into as many equal spaces 
as there are teeth, and lay off the thickness of the teeth equal to one-half of the pitch, 
and draw the outline of the teeth as clearly shown in the section. 

It now remains to determine the contour and its limits at the edge or end of the 
teeth, which may be done as follows : From where a h intersects the outside, the pitch 
and the root cylinders of the worm, project to the corresponding helices as shown by 
the dotted outline in the front view. This will give a foreshortened outline of the 
worm thread space as shown. Produce a h to the axis of the wheel as at g'', draw 
a" g parallel to a g"", and project c to c^'; also the intersection of the outside and root 
cylinder with a h, which will give the limit of the worm thread space above and 
below the pitch line c c". Then lay off the outline of the section in the same relation 
to C g as the outline of the section in the front view is to p d ; this will be a parallel 
view of the worm thread space. Transfer this view to an extra piece of paper as 
shown at p g^, lay off on the pitch line equal divisions each side of p, also the same 
divisions on the axis of the wheel the same number of times each side of g^. With 
g^ p as a radius and the points on the axis each side of g"" as centres, draw arcs tangent 
to each point on the pitch line as shown, which will represent the pitch circle in 
different positions as it rolls on the pitch line of the worm. Then on a piece of 
tracing cloth draw the pitch line of the worm and perpendicular to it a line equal to 
p g^ ; with p g'' as a radius and g^ as a centre draw an arc of the pitch circle ; through 
p on this arc of the pitch circle lay off divisions equal to those on the drawing. 
Place the tracing on the drawing so that p and g^ on the tracing coincides with p g'' 
on the drawing, fasten with thumb tacks and trace the section, remove the tacks 
and revolve the tracing until the pitch circle on the tracing coincides with the next 
arc of the pitch circle on the drawing, and the next point to p on the tracing coincides 
with the next point on the pitch line, fasten with the thumb tacks and trace the 
section again. Repeat this four or five times in each direction, then the envelope of 
these tracings will be the contour of the wheel tooth on the parallel plane, as seen in 
the parallel view at c''. Find centres from which this outline can be drawn and 
transfer it from the tracing to the parallel view in the same relation to c'' g as it is to 
c'' g on the tracing. By projecting from the sectional view as shown at the top, the 
radii of the circles which limit the curves of the teeth at the outside and bottom will 
be obtained. These circles having been drawn, lay off on these circles and the pitch 
circle the points through which to draw the left outline of a tooth in the same 

60 



relation to p d as they are to a' g in the parallel view, then by making the width at 
the outside the pitch and the bottom equal to the width at the corresponding circles 
in the parallel view will give the points through which to draw the other side of the 
tooth. Then by drawing these outlines in the same relation to all the pitch points 
the front side of the teeth are finished. The outlines of the rear side are found in 
the same manner, by first finding an outline of the worm thread space on the rear 
side of the same space. In drawing these arcs upon which to place the tracing cloth. 



P lat e- ^ 7 - 




the longer they are the better for accuracy. This is a very difiicult operation, there- 
fore great care should be excerised. The outside diameter of the wheel may be 
determined as follows : The radius of the root cylinder plus the clearance multiplied 
by the versed sine of 30°, and this product multiplied by 2, then this last product 
added to what would be the outside diameter of a spur wheel having the same number 
of teeth and pitch, will give outside diameter of the worm wheel. Versed sin. of 



30^ 



13397 



61 



Plate 28. 

The statement has been made that the path of contact of involute teeth was along 
the line of action. It will be well to give this more attention now, as the pu^il is 
better acquainted with the subject, and therefore better prepared for further explana- 
tion. Draw the pitch circles of two wheels, and lay off the line of action. Then 
draw the base circles through a and b, and draw the addendum circle of the wheel 
e^ through a, and the addendum circle of the wheel d"" through b. 

lyCt the flexible band f "^ b a a'' be made fast to the base circles at f ' and a^, then 
if e" be revolved as indicated by the arrow, the flexible band will be wound on e"" and 
unwound from d'', and at the same time revolve d'' as indicated by the arrow. Then 
while the wheels are revolving, a point fastened to the band at a will move along a b, 
and during this movement will generate the involutes a g and a c. It should be 
readily seen that the teeth must be in contact along the line of action. 

It will also be seen ; that the contact begins at a and ends at b, which is the 
greatest length of the path of contact, and as the addendum circles pass through a 
and b the full length of contact is made use of. 



62 



FMq. ?r^ 




SCREW CUTTING. 

Rule. — Divide pitch of screw to be cut, b}^ pitch of lead screw. Pitch of a screw 
is the distance it travels in one revolution. The lead screw is that screw which moves 
the lathe carriage along the shears of the lathe, and the movement of the carriage 
per revolution of the work is regulated by gear wheels on the end of the lathe. 

To cut 4 threads per inch with a lead screw of 5 threads per inch : Four threads 

per inch is equal to %'' pitch ; the lead screw 5 threads per inch equals |^ pitch. Then 

I I 1,5 5 = stud wheel. 

4*5 4 I 4 =:^ lead screw wheel. 

But there are no wheels with such a small number of teeth, and to obtain a suitable 

number of teeth, we may multiply { by any number as 10; then 

S , SO = stud wheel. 

— X 10 ^=^ 

4 40 = lead screw wheel. 

If the thread to be cut is a certain number of threads and a fraction of a thread in one 

inch, as 5^ threads per inch, it will be 5^ equals -^^, which must be divided into i 

inch to obtain the pitch, then 

4 21 21 

pitch or 21 threads in 4 inches, and is similar to the former example. )/'' pitch is 4 
threads in one inch, the numerator of the fraction always denoting the number of 
inches, and the denominator the number of threads, in that number of inches. Then 

4 5 20 = stud wheel. 

21 I 21 = lead screw wheel. 
If the thread to be cut should be one thread in more than one inch, as one thread in 
1% inches, then 1% .= \, which is the pitch, and 

7^i^7^5^35 = stud wheel ; ^j. 35 v^ 140 = stud. 

45 4 I 4 = lead screw wheel ; 4 16 = lead screw. 

But there is a great difference between the two wheels, which in most cases will make 
it necessary to compound, which may be done as follows : Divide ^ by any numbers 
which will divide them without a remainder, as 

35^5 = 7 
4 H- 2 = 2* 

Then f , ^ are the new wheels, and 

-5xio = 5°and2x8=5|, 

2 20 2 16 

or 1^, f f will do the same work as y/-, and will go on the lathe in this manner : Put 
50 on the stud gearing in 20, then 56 along side of it, gearing in 16 on the lead screw. 
56 and 20 must be keyed to a sleeve which is made to turn on the intermediate stud. 
The numerator and the denominator must be multiplied by the same number. Any 
number that may be necessary to obtain a set of wheels from the wheels there may be 
to select from may be used. Whatever may be the pitch of the screw to be cut, or 
the pitch of the lead screw, the wheels are obtained as above. 

Threads are quite frequently irregular, as 4^ threads in 3^ inches. 

\ 4 ^8 8 ' 

64 



-\^- is the length which contains 4^ or-^- threads. The length divided by the number 
of threads equals pitch, as 

8 4 8 19 152 

this being large may be reduced without changing its value, 

116 29 . , 

— = = pitch, 
152 38 

which, divided by pitch of lead screw, equals wheels. I^et pitch of lead screw be 4 
threads per inch, equals % pitch, then 

29 4 ^ £i_6 

38 I 38 • 

116^-^1=29^^ 58 J 4,, 40 

-— ~ -^- X 2 = ^~ and ^ X 10 = -- . 

38 ^- 2 = 19 38 2 20 

Then fg and ff are the wheels that will cut 4^ threads in 3^ inches. 

To prove a set of wheels, divide the product of the teeth of the driven wheels b}^ 

the pitch of the lead screw, then this quotient divided by the product of the teeth of 

the driving wheels will give the number of threads per inch. Or, if there are only two 

wheels, then the number of teeth in the lead screw wheel divided by pitch of lead 

screw, and this quotient divided by number of teeth in stud wheel, will give the 

number of threads per inch. 



65 



Table No. 1. 
20° LINE OF ACTION, CIRCULAR PITCH. 

Note : — This table can be used only for 20° line of action. Multiply tabular 
numbers by circular pitch to obtain the different radii. 
For 20° line of action : 

Radius of face curve = .34202 X r + (.2125 p). 
Radius of flank curve = .04125 X p X n. 

r = radius of pitch circle. 

p = circular pitch. 

n = number of teeth. 



Column No. 1. 
No. of teeth. 


Column No. 2. 
Radius of face. 


Column No. 3. 
Radius of flank. 


Column No. 4. 

Radius of circle 

at centre. 


Column No. 5, 
Radius of fillet. 


12 


.86568 


•495 


.2 


•15 


. 13 


.92 


.53625 


.15 




•15 


14 


- -97452 


•5775 


•125 




.125 


15 


1.02896 


.61875 


Radial 




.125 


16 


I. 1034 


.66 


( ( 




125 


17 


1. 1378 


.70125 


(( 




125 


18 


I. 19224 


•7425 


( ( 




125 


20 


I.3QI14 


.825 


Curve 




I 


22 


I. 4 1004 


.9075 


t < 




I 


24 


1. 5188 


-99 


< < 




075 


26 


1.6277 


1.0725 


i< 




075 


30 


1.8454 


1-2375 


( ( 




05 


40 


2.38978 


1.65 


( ( 




0375 


50 


2.934 


2.062 


( ( 




0375 


70 


4.0227 


2.887 


( ( 




0375 


100 


5-6557 


4.125 


( ( 




0375 


150 


8-3773 


6.187 


(( 




0375 


300 


16.542 


12.375 


( ( 




0375 


Rack 











66 



Table No. 2. 
20° LINE OF ACTION, DIAMETRAL PITCH. 

Note : — This table can be used only for 20° line of action. Divide tabular num- 
bers by diametral pitch to obtain the radii. 



Column No. 1. 
No. of teeth. 


Column No. 2. 
Kadius of face. 


Column No. 3. 
Radius of flank. 


Column No. 4. p , 
Radius of circle .V^i^ 
at centre. ^^'^' 


mn No. 5. 

as of fillet. 


12 


2.7193 


I-555I ^ 


.3148 


4712 


13 


2.8902 


1.6845 


•2356 


4712 


14 


3.0614 


1. 8142 


.1963 


3927 


15 


3-2323 


1-9437 


Radial 


-3927 


16 


3.4664 


2.0734 


( i 


3927 


17 


3-5745 


2.2028 


( < 


3927 


18 


3-7454 


2.3326 


( ( 


3927 


20 


4.0875 


2.5918 


Curve 


3141 


22 


4.4296 


2.8510 


( ( 


3141 


24 


4-7714 


3. 1 100 


( < 


2356 


26 


5-II35 


3-3693 


I ( 


2356 


30 


5-7975 


3-8877 


( ( 


1570 


40 


7.5074 


5-1836 


(( 


II78 


50 


9.2174 


6.4779 


( ( 


II78 


70 


12.6377 


9.0697 


( ( 


II78 


100 


17.7679 


12.9591 


<( 


II78 


150 


26.3181 


19.4370 


({ 


II78 


300 


51-9683 


38.8773 


<( 


II78 


Rack 











67 



Table No. 3. 
14i° LINE OF ACTION, CIRCULAR PITCH. 

Note: — This table can be used only for i4>^° line of action. Multiply tabular 
numbers by circular pitch to obtain the different radii. 
For 1 4/^° line of action : 

Radius of face curve = .25038 X r -f (.2125 p). 
Radius of flank curve = .033 X p X n. 

r = radius of pitch circle. 

p = circular pitch. 

n = number of teeth . 



Column No. 1. 
No. of teeth. 


Column No. 2. 
Kadius of face. 


Column No. 3 
Radius of flank. 


Column No. 4. ^ . 
Radius of circle ^^ ^,'3 
at centre. ^^'' 


mn No. 5. 
js of 'fillet. 


12 


.6906 


.396 


•23125 


175 


^3 


•7305 


.429 


.1982 


.1708 


14 


•7703 


.462 


.1652 


1666 


15 


.8102 


•495 


.1321 


1625 


16 


.8500 


.528 


.0991 


1583 


17 


.8899 


.561 


.0661 


1542 


18 


.9297 


■594 


•0330 


1500 


20 


1.0094 


.660 


Radial 


1417 


22 


1. 0891 


.726 


( < 


1333 


24 


I. 1688 


.792 


< < 


1250 


26 


1.2485 


.858 


( ( 


1 167 


30 


1.4079 


.990 


(i 


1000 


40 


1.8064 


1.320 


Curve 


0583 


50 


2.2048 


1.650 


( ( 


0500 


70 


3.0018 


2.310 


( ( 


0500 


100 


4.1972 


3-300 


t< 


0500 


150 


6.1896 


4-950 


( ( 


0500 


300 


6.1896 


9.900 


( ( 


0500 


Rack 











68 



Table No. 4. 
14J° LINE OF ACTION, DIAMETRAL PITCH. 

Note: — This table can be used only for 14)^° line of action. Divide tabular 
numbers by diametral pitch to obtain the radii. 



Column No. 1. 
No. of teeth. 


Column No. 2. 
Radius of face. 


Column No. 3. 
Radius of flank. 


Column No. 4. ^ , 
Radius of circle j^ ^-^ 
at centre. ^^^ 


imn No. 5. 
lus of fillet. 


12 


2.1696 


1. 2441 


.7264 


54978 


13 


2.2949 


1.3477 




6227 


53658 


14 


2.4200 


I-4514 




5189 


52339 


15 


2.5453 


I-5551 




4151 


51051 


16 


2.6704 


1.6588 




3114 


49731 


17 


2.7957 


1.7624 




2076 


48443 


18 


2.9207 


1. 8661 




.1038 


47124 


20 


3.1711 


2.0734 


Radial 


44516 


22 


3-4215 


2.2808 


< ( 


41877 


24 


3.6719 


2.4881 


i i 


39270 


26 


4.0223 


2.6955 


(( 


36662 


30 


4.4230 


3. 1 102 


( i 


31416 


40 


5^6750 


4.1469 


Curve 


1835 


50 


6.9266 


5-1836 


t i 


15708 


70 


9.4304 


7-2571 


( ( 


15708 


100 


13-1859 


10.3672 


( ( 


15708 


150 


19.4452 


15-5509 


1 ( 


15708 


300 


19.4452 


31.1018 


( ( 


15708 


Rack 











69 



Table No. 5. 

FRANKLIN INSTITUTE OR U. S. STANDARD BOLT AND 

NUT PROPORTIONS. 












ic 


^ 
-M 




1 








as 


10 


tH 


■tlJ . 


C3 V 




o 

u 




rH 
If . 


II V- 

03 '"'irH 

4) 4) 


^ X 


If 


(D 





0^ 

ft 

00- 


CS T-l 


'S 


^1 


03 c3 




-4J 

1 a 


1 


© 

a 


ciS 


S 1 

o 


,13 


Is 




2 -S 

•l-H 




1// 


20 


.185" 


.0062" 


7 '^ 


.505" 


3 ff 

16 


9 " 
TF 


3 " 
"32 


5 'f 
T6 


1 8 


.240" 


.0074" 


3 2 


.613" 


1'/ 
4 


11'/ 
T15" 


1" 


3'/ 
■g" 


i6 


.294" 


.0078'' 


5/^ 


.722" 


5 f' 
T6 


13'/ 
T6 


1" 
8 


tV 


H 


•344" 


.0089'' 


11" 


.830" 


3// 


15" 
16^ 


5 " 
32 


i" ' 


13 


.400" 


.0096'' 


13'^ 
T6 


•938" 


7 '/ 
16 


T 1 '^ 


5 " 
32 


A" 


12 


•454" 


.0104" 


2 9'^ 
32 


1.046" 


2 


T 3 " 
I16 


3 " 
T6 


i" 


II 


•507" 


.0113'' 


l" 


..154" 


tV 


T 5 " 

Ire 


3 " 
1 6 


3" 


lO 


.620" 


.0125'' 


^16 


I.37I" 


11'/ 
T6 


T 9 " 
lye 


7 " 
32 


7^' 
8 


9 


•73^" 


.0138'' 




1.587" 


13/' 

T6 


t13" 
^16 


7 " 
32 


// 


8 


•837" 


.0156'' 


T 9 " 


1.804" 


IS/' 

re 


2tf 


1" 

4 


li" 


7 


.940" 


.0178'' 


I4 


2.020" 


T 1 '' 


5 " 
2T6 


1" 
4 


t1" 
^4 


7 


1.065" 


.0178'' 


iH" 


2.237" 


T 3 /' 

^16 


^ 9 " 


5 " 
1 6 


If" 


6 


1. 160" 


.0208" 


2i" 


2.453" 


T 5 // 


^13" 
^1 e 


5 " 
TF 


^2 


6 


1.284" 


.0208'' 


2T6 


2.670" 


I '^/' 


3tV' 


3" 
8" 


5ff 


5i 


1.389" 


.0227'^ 


2i'' 

^^2 


2.886" 


T 9 " 
^16 


3i6 


3" 

8 




5 


1.491" 


.0250" 


2H" 


3.103" 


IT6 


^ 9 " 

3te 


7 " 
TF 


If 


5 


1. 616" 


.0250'' 


21" 


3-319" 


lie 


^13" 

3 IF 


1" 

2 


2" 


4* 


1. 712" 


.0277" 


3t6 


3.536" 


^16 


4tV' 


9 " 
T6 


2i" 


4i 


1.962" 


.0277^' 


016 


3.969" 


-^16 


4iV' 


5" 
8 


^r 


4 


2.176" 


.0312" 


.-,13'/ 
016 


4.402" 


^16 


5t6 


11" 
TF 


2|" 


4 


2.426" 


.0312'' 


^ 3 '/ 

4T6 


4.835" 


^IF 


C 9 " 

ore" 


3" 

4 .- 


3" 


3i 


2.629" 


•0357" 


4T6 


5-268" 


2t6 


^16 


13" 

16 



These are finished sizes. Add -§■'' for rough sizes instead of j^^\ 



70 



DECIMAL EQUIVALENTS. 



8ths. 


A = 


=•21875 


ii= -265625 


/s = .125 


9 — 
■32 - 


=■28125 


if -.296875 


'A = -250 


11 — 

S2 - 


= -34375 


|i = . 328125 


^8 = .375 


13 _ 
32 - 


= -40625 


If = -359375 


^ = .500 


15 — 
^2^ - 
1 


= -46875 


li = -390625 


5/8 = .625 


H = 


= -53125 


11= .421875 


H = -750 


if = 


= -59375 


If = -453125 


n = -875 


21 _ 


= -65625 


li = -484375 




23 _ 
32 ~ 


= .71875 


11= -515625 


leths. 


25 — 

32 — 

27 _ 
32 - 


= -78125 
= -84375 


If = .546875 

IJ-. 578125 


tV = -0625 


29 — 
32 - 


= .90625 


If = -609375 


fV = .i875 


31 — 
^2" — 


= -96875 


tl = .640625 


T6=-3I25 






If =-671875 


T6 = -4375 

A = -5625 


64ths. 


If = -703125 
If =-734375 


H = -6875 
Ii = -8i25 


1 

"54 

3 

64 


-015625 
-046875 


n = -765625 
|i=. 796875 


H = -9375 


5 

6 4 


.078125 


|i = . 828125 


32nds. 


9 


-109375 
. 140625 


6i = '859375 
IJ = .890625 


3\= -03125 


11 

6 4 


•171875 


11 = .921875 


A = -09375 


la — 

64 


.203125 


fi = -953125 


A = -15625 


15 

6 4 


•234375 


If = .984375 



71 



